Functions of a Complex Variable

Date: 10/05/2002 at 11:28:34
From: Jay
Subject: Functions of a complex variable

Show that w=f(z*), where z*=x-iy and f is a differentiable function, 
is not an analytic function of z.

Date: 11/11/2002 at 11:43:14
From: Doctor Nitrogen
Subject: Re: Functions of a complex variable

Hi, Jay:

This gets pretty involved, so I hope you have time to read carefully.

The only proofs I could find somewhat related to my argument here can 
be found on page 71, problem 2, and page 32, problem 5,in:

   _Theory and Problems of Complex Variables_,
   Murray Spiegel, 
   McGraw Hill, 1964, Schaum's Outlines, ISBN #07-060230-1.

Do you recall those delta(y)/delta(x) equations you had to use when 
you first learned to calculate a derivative in calculus? That is, if 
you recall, you had to compute a limit of something like

   [f(x + delta(x)) - f(x)]/delta(x), 

to find the derivative of something like

   f(x) = x^2.

To prove that f(z*) fails to be analytic, you will have to do 
something similar.

First of all, let 

   w = f(z*) 

     = ((u(x, y) + iv(x, y))*.

By the definition of derivative in Complex Variable Theory:

   (d/dz*)f(z*)

 = Limit         [f((z + delta(z))*) - f(z*)]/(delta(z*)).
    delta(z*)->0
 
First, consider what happens when delta(y) = 0, lim delta(x)-> 0:

When (delta(y) = 0),


    Limit         [f((z + delta(z))*) - f(z*)]/(delta(z*))
     delta(x)->0

  = Limit          {  [u(x + delta(x), y) - iv(x + delta(x), y)] 
     delta(x)->0              

                    - [u(x, y) - iv(x, y)] } / delta(x) 


  = Limit          {   u(x + delta(x), y) - u(x, y) 
     delta(x)->0

                     - i[v(x + delta(x), y) + iv(x, y)] } /delta(x) 


  = Limit          { u(x + delta(x), y) - u(x, y) } / delta(x) 
     delta(x)->0

                 - [i(v(x + delta(x), y) - iv(x, y)] / delta(x)


[A]              = &u/&x - i&v/&x,

where "&this/&that" denotes "the partial derivative of this with 
respect to that." 

Now consider what happens when delta(x) = 0 and lim delta(y) -> 0:

When (delta(x) = 0),

    Limit          [f(z + delta(z))*) - f(z*)]/(-i delta(y))
     delta(y)->0
  
  = Limit         {  [u(x, y + delta(y)) - iv(x, y + delta(y))] 
     delta(y_->0                 

                   - [u(x, y) - iv(x, y)] } / (-i delta(y)) 

 
  = Limit         {  [u(x, y + delta(y)) - u(x, y)] 
     delta(y)->0

                   - [iv(x, y + delta(y) + iv(x, y)]} / (-i delta(y))

  
  = Limit         {  [u(x, y + delta(y)) - u(x, y)] 
     delta(y)

                   + [iv(x, y + delta(y) - iv(x, y)] } / (i delta(y))

 [B]         = &u/&y + i&v/&y.


Now recall that for a complex valued function to be analytic, a 
necessary and sufficient condition is for the Cauchy-Riemann equations 
to be satisfied, that is:

 
[C]              &u/&x + i&v/&x = &v/&y - i&u/&y, 

         or
                 &u/&x = &v/&y, 

                 &u/&y = -&v/&x.

But compare what you must have for f(z*) to be analytic from [A] and 
[B]:


  [A]             &u/&x - i&v/&x


  [B]             &u/&y + i&v/&y,

or it must be that

  [D]             &u/&x = &u/&y,  

                  &v/&y = -&v/&x.


Clearly, [C] and [D] are not the same conditions, are they? But it is 
known that [C], or the Cauchy-Riemann equations, must be satisfied, 
so [C] is true, meaning the equations in [D] cannot be true, so f(z*) 
can't be analytic. 
  
You also might ask yourself, as another way to prove f(z*) fails to 
be analytic, (I did not investigate this myself, but you might want 
to) whether both 

   Re[f(z*)] and Im[f(z*)]

are harmonic functions, since every analytic function has two harmonic 
functions associated with it, that is, 

   f(z) = u(x, y) + iv(x, y) 

is an analytic function provided u(x, y) and v(x, y) are both 
harmonic. If you can show Re[f(z*)] and Im[f(z*)] are not harmonic, 
it would follow they are not analytic. 

I hope this helped answer the questions you had concerning your 
mathematics problem. You are welcome to return to The Math Forum and 
Doctor Math whenever you have any math-related questions.

- Doctor Nitrogen, The Math Forum
  http://mathforum.org/dr.math/